# Homework Help: Help with capacitors

1. Jun 20, 2010

### TsAmE

1. The problem statement, all variables and given/known data

I dont get how this circuit is a short-circuit as there is a resistor connected in this circuit, which is opposing the flow of current.

How can both terminals of the capacitor be connected to earth, shouldnt one be connected to the positive and the other to the negative (in order for a current to flow)?

2. Relevant equations

None.

3. The attempt at a solution

n/a.

#### Attached Files:

• ###### Capacitor.png
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2. Jun 20, 2010

### n.karthick

hey, here the capacitor is going to discharge its charge (if there is any).
It is like when you release a compressed spring. The spring will tend to come back to its original shape isn't it? The charged capacitor can be thought of as a compressed spring. Can you tell the significance of resistance value in discharging of capacitor, like what will be the discharging time if you have large resistance? will the discharging time change?

3. Jun 20, 2010

### poor mystic

Hi TsAmE
it looks like they want you to think of a previously charged capacitor and how it discharges when connected to a short circuit.

Capacitors work by storing electric charge. If you were to connect a capacitor to a battery for a moment, electrons would migrate from the battery onto one 'plate' of the capacitor.
Then you could connect the capacitor to a load (usually symbolised by a resistor) and for a moment there would be some electric current flowing in the load resistor.

A charge can be kept on a capacitor for quite long periods. Any imbalance of charges on the capacitor would soon be removed by this circuit, with the current in the resistor at any time being exactly proportional to the charge left on the capacitor.

4. Jun 20, 2010

### TsAmE

I know that the larger the resistance, the larger the discharging time (Time constant RC).

Sorry but I still dont get why it is a short-circuit if a resistor is contained in the circuit.

Since a capacitor is like a battery which stores charge, why isnt it connected with the positive terminal as the supply and the negative terminal as earth? (Refer to new attachment).

#### Attached Files:

• ###### Capacitor 2.png
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112
5. Jun 20, 2010

### Integral

Staff Emeritus
The short circuit is just referring to the red section. Perhaps in a different problem they but some sort of load, eg a resistance, in that place. Both terminals of the cap are not connected to directly to ground, the positive side (as shown in your pic) is connected to the resister. Current must flow through it to reach ground.

6. Jun 20, 2010

### TsAmE

If that is the case why is it that the discharging current starts and ends at the terminals of the short-circuit

I dont see how the conventional current flows from the positive to negative in this circuit.

7. Jun 20, 2010

### Integral

Staff Emeritus
I don't understand what you are saying. Why do you think that the current "starts and ends", what ever that means, with the short?

8. Jun 21, 2010

### poor mystic

I remember similar levels of confusion in early classes. The thing to remember is that the whole mechanism is just as simple as a bucket of water which suddenly develops a hole in the bottom. If it seems more complicated than that, you're still confused.

The 'water' is the electric charge imbalance on the capacitor. This charge will move through whatever conducting medium is available to get to the other side of the capacitor and balance the charge.

The smaller the hole in the bucket, the higher the resistance the hole has to water flowing through it, and the slower the discharge. For the discharging capacitor a similar property of the resistor in the circuit slows charge movement.

By the way, don't read too much into the term 'short circuit'. It is confusing, but in this case it doesn't mean a 'dead short' or blowing fuses.

Last edited: Jun 21, 2010
9. Jun 21, 2010

### kntsy

1.short circuit refers to the colored area only.
2.There is no need to connect the charged capacitor to a battery to discharge it; Connecting both plates by non-insulators(wires/Earth) enables the flow of electrons and discharge occurs.

10. Jun 21, 2010

### TsAmE

Yeah I understand the way a capacitor operates, but it is just the way in which the current flows which is confusing me.

I understand this, but if the flow of electrons are enabled and discharge occurs, where would the electrons start (i.e. negative terminal?) and end (i.e. positive terminal)?

11. Jun 21, 2010

### RajChakrabrty

TsAme,

I am considering,the 1st picture you have attached.
one has rightly said that short circuit area is the red part only,so concentrate on remaining part.
But, there is another confusion ,anyone could have mentioned, that switch is not closed.
if switch will remain open then Discharging of Capacitor is not possible.

here we can make one anticipation , based on the given ckt.........

** the short ckt part contains an Inductor----
as, when the switch will be closed the inductor will be considered as short ckt for t=0+
0+(zero +) means the time is just after closing of switch.
this may be the reason for showing the short ckt there.

[you may know,
the inductor acts as ahort ckt and capacitor as open ckt at t =0+
and
inductor as open ckt and capacitor as short ckt for t=infinite]

this is the answer of short ckt issue.

now, the current flowing in the ckt and the GND problem.
I have seen that some one has discussed beautifully with water flowing concept.
so,
if you have still any doubt,
Best of Luck.

12. Jun 21, 2010

### TsAmE

Thanks. I understand the water flowing concept, but what I think is confusing me is that I am thinking of this circuit in terms of any circuit, where one wire must be connected to + and the other to - (thus producing a conventional current), but here both wires are connected to earth (where the dot is). This is why I cant see how a current can flow.

13. Jun 21, 2010

### poor mystic

Ah well. In this case, the + and - charges have been forced to take up residence on the capacitor. Any conduction path will do, that charge must be balanced.
But notice! the capacitor works like a battery, dry cell or whatever... it has a negative and a positive terminal, and current certainly flows in a loop from the device by one wire and back to the device by the other wire.

Last edited: Jun 21, 2010
14. Jun 21, 2010

### poor mystic

Take a dry cell and a small lamp. Connect the lamp to the cell, check that the lamp lights.
Connect one side of the dry cell to earth, the other side to the lamp and through the lamp to earth. The lamp lights.
The Earth can be a conductor as well as a terminal.

15. Jun 21, 2010

### RajChakrabrty

discussion by Poor mystic is wonderful.

16. Jun 21, 2010

### poor mystic

poor poor mystic suspects irony

17. Jun 22, 2010

### TsAmE

Thanks! I understand now :). Why is it that in the diagram the discharging current doesnt point from the + terminal of the capacitor to the - ?

18. Jun 22, 2010

### Zryn

The blue loop indicating the polarity of the current is a poor illustration of where the electrons move from and to. They do not start at ground, and move through the capacitor to the switch but rather, will travel from the top (+) of the capacitor to the bottom (-) along the path of least resistance, through whatever medium necessary, providing they have sufficient force (voltage) to do so.

19. Jun 23, 2010

### TsAmE

I see. Another doubt which I have had is how the capacitor charges up to the supply rail, even if a resistor is present. Say in the diagram you used a +13V supply rail and the capacitor is charging. A certain amount of voltage will be dropped across the resistor (say 5V is across R). Now what I want to know is why is it that the capacitor can still charge up to +13V and not Vc (13V - 5V = 6V)? As the 5V was lost when the current traveled through R.

#### Attached Files:

• ###### Capacitor 3.png
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Last edited: Jun 23, 2010
20. Jun 23, 2010

### Zryn

What do you know about capacitors charging? Click on the 'capacitor' link in your post and scroll down to 'Inverse exponential rate of charging'.

Capacitors are non-linear (exponential) when they are charging / discharging (over time). When they reach equillibrium (always 5 RC time constants worth of charging / discharging; though like an exponential they can never truly reach 100% charge or discharge) you can use linear equations such as Ohms Law.

* If the switch in your diagram opens faster than 5 RC time constants, the capacitor will always be charging or discharging and so you will not be able to use Ohms Law to determine the current in the circuit.

21. Jun 27, 2010

### TsAmE

Yeah I understand this, but still dont get why it (almost) charges up to the supply voltage, despite there being a resistor in the way.

22. Jun 27, 2010

### Zryn

As long as the capacitor's charge is not equal to the source potential it will continue to charge.

As long as the capacitor is charging, the voltage dissipated across the resistance will be decreasing over time, the same as the capacitor voltage will be increasing over time, and the value of the circuit current will change over time, as per the non-linear exponential charging equations.

Eventually, when time reaches infinity, the capacitor will hold the same voltage as the source, and the current will be 0A, and therefore by Ohms law, the voltage drop across the resistor will be 0A*R = 0V.

This all happens because when time has not yet reached infinity, the circuit has not yet reached a stable equilibrium yet, much like a spring that is trying to uncoil, but takes infinity time to do so, it will continue trying to uncoil until it reaches its stable equillibrium.

23. Jun 27, 2010

### vk6kro

The resistor does not have a fixed voltage across it.

The capacitor starts off with zero voltage on it. So all the supply voltage must be across the resistor. (so you can calculate the current. I = E / R).
With a current flowing into it, the capacitor starts to charge up and get a voltage across it.
This means the resistor has less voltage across it, so the current decreases.

If the supply is 12 volts, and the capacitor has charged to 6 volts, there must be 6 volts across the resistor.

As the capacitor charges more and more, the current keeps dropping. If you wait long enough, the capacitor voltage will get very close to the supply voltage. Depending on the component values, this may only need a millisecond or so, or it could take hours with big capacitors and resistors.

24. Jun 28, 2010

### TsAmE

I see, so voltage division actually occurs between R and the capacitor while it is busy charging? (as there is a certain resistance across the capacitor?)

Also when the capacitor has fully charged and there is no voltage drop across the resistor, could it be said that the capacitor has an infinite resistance? (as all of the supply voltage is dropped across it?)

25. Jun 28, 2010

### Zryn

A capacitor is comprised of two conductive plates separated by a dielectric, (air, ceramic polymer, could be almost anything non conductive) as the capacitance of a capacitor is related to the distance between the plates, the stronger the dielectric strength, the larger a voltage it can separate without failing and creating a short circuit.

This means when fully charged the capacitor looks like an open circuit, i.e. infinite resistance because of this dielectric separating the two conducting plates.

When the capacitor is charging, current doesnt flow >>through<< it, like current flows through a resistor (creating a voltage drop), since that would mean the dielectric has failed and the capacitor would be a short circuit. Rather, charged particles flow onto the positive plate, and can then flow out of the positive plate when the capacitor is discharging.

Ideal capacitors have no resistance across them like you imply (in a DC circuit), but yes, there is a voltage division between the resistor and the capacitor (since kirchoffs voltage law - based on conservation of energy - still has to be adhered to, and there is only two components to divide the voltage between).